3.664 \(\int \frac{x^3 \sin ^{-1}(x)}{(1-x^2)^{3/2}} \, dx\)

Optimal. Leaf size=36 \[ \sqrt{1-x^2} \sin ^{-1}(x)+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}-x-\tanh ^{-1}(x) \]

[Out]

-x + ArcSin[x]/Sqrt[1 - x^2] + Sqrt[1 - x^2]*ArcSin[x] - ArcTanh[x]

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Rubi [A]  time = 0.0692451, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {266, 43, 4689, 388, 206} \[ \sqrt{1-x^2} \sin ^{-1}(x)+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}-x-\tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcSin[x])/(1 - x^2)^(3/2),x]

[Out]

-x + ArcSin[x]/Sqrt[1 - x^2] + Sqrt[1 - x^2]*ArcSin[x] - ArcTanh[x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 4689

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSin[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 - c
^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2
, 0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^3 \sin ^{-1}(x)}{\left (1-x^2\right )^{3/2}} \, dx &=\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}+\sqrt{1-x^2} \sin ^{-1}(x)-\int \frac{2-x^2}{1-x^2} \, dx\\ &=-x+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}+\sqrt{1-x^2} \sin ^{-1}(x)-\int \frac{1}{1-x^2} \, dx\\ &=-x+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}+\sqrt{1-x^2} \sin ^{-1}(x)-\tanh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.064117, size = 40, normalized size = 1.11 \[ \frac{1}{2} \left (-\frac{2 \left (x^2-2\right ) \sin ^{-1}(x)}{\sqrt{1-x^2}}-2 x+\log (1-x)-\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcSin[x])/(1 - x^2)^(3/2),x]

[Out]

(-2*x - (2*(-2 + x^2)*ArcSin[x])/Sqrt[1 - x^2] + Log[1 - x] - Log[1 + x])/2

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Maple [A]  time = 0.219, size = 61, normalized size = 1.7 \begin{align*} -x+\arcsin \left ( x \right ) \sqrt{-{x}^{2}+1}-{\frac{\arcsin \left ( x \right ) }{{x}^{2}-1}\sqrt{-{x}^{2}+1}}-\ln \left ({\frac{1}{\sqrt{-{x}^{2}+1}}}+{x{\frac{1}{\sqrt{-{x}^{2}+1}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(x)/(-x^2+1)^(3/2),x)

[Out]

-x+arcsin(x)*(-x^2+1)^(1/2)-(-x^2+1)^(1/2)/(x^2-1)*arcsin(x)-ln(1/(-x^2+1)^(1/2)+x/(-x^2+1)^(1/2))

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Maxima [A]  time = 1.44542, size = 61, normalized size = 1.69 \begin{align*} -{\left (\frac{x^{2}}{\sqrt{-x^{2} + 1}} - \frac{2}{\sqrt{-x^{2} + 1}}\right )} \arcsin \left (x\right ) - x - \frac{1}{2} \, \log \left (x + 1\right ) + \frac{1}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^2+1)^(3/2),x, algorithm="maxima")

[Out]

-(x^2/sqrt(-x^2 + 1) - 2/sqrt(-x^2 + 1))*arcsin(x) - x - 1/2*log(x + 1) + 1/2*log(x - 1)

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Fricas [A]  time = 2.50162, size = 155, normalized size = 4.31 \begin{align*} -\frac{2 \, x^{3} - 2 \,{\left (x^{2} - 2\right )} \sqrt{-x^{2} + 1} \arcsin \left (x\right ) +{\left (x^{2} - 1\right )} \log \left (x + 1\right ) -{\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 2 \, x}{2 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^2+1)^(3/2),x, algorithm="fricas")

[Out]

-1/2*(2*x^3 - 2*(x^2 - 2)*sqrt(-x^2 + 1)*arcsin(x) + (x^2 - 1)*log(x + 1) - (x^2 - 1)*log(x - 1) - 2*x)/(x^2 -
1)

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Sympy [A]  time = 15.3001, size = 37, normalized size = 1.03 \begin{align*} - x - \left (- \sqrt{1 - x^{2}} - \frac{1}{\sqrt{1 - x^{2}}}\right ) \operatorname{asin}{\left (x \right )} + \frac{\log{\left (x - 1 \right )}}{2} - \frac{\log{\left (x + 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asin(x)/(-x**2+1)**(3/2),x)

[Out]

-x - (-sqrt(1 - x**2) - 1/sqrt(1 - x**2))*asin(x) + log(x - 1)/2 - log(x + 1)/2

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Giac [A]  time = 1.08882, size = 54, normalized size = 1.5 \begin{align*}{\left (\sqrt{-x^{2} + 1} + \frac{1}{\sqrt{-x^{2} + 1}}\right )} \arcsin \left (x\right ) - x - \frac{1}{2} \, \log \left ({\left | x + 1 \right |}\right ) + \frac{1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^2+1)^(3/2),x, algorithm="giac")

[Out]

(sqrt(-x^2 + 1) + 1/sqrt(-x^2 + 1))*arcsin(x) - x - 1/2*log(abs(x + 1)) + 1/2*log(abs(x - 1))